Question

The daily consumption of electric power (in millions KW-Hours) is a random variable having the p.d.f f(x)= {█(1/9 xe^(-x/3) ,x>0@0 ,x≤0)┤ if the total production is 12 millions KW- Hours. Determine the probability there is power cut (shortage) on any given day.

Answer 1

Concept

In probability theory, a probability density function is a function whose value at any given sample in the sample space can be interpreted as providing a relative likelihood that the value of the random variable would be near to that sample.

Given

daily consumption of electric power (in millions KW-Hours) is a random variable having the p.d.f f(x)= {(1/9 xe^(-x/3) ,x>0)

Find

we have to determine the probability there is power cut (shortage) on any given day.

Solution

The probability density function is given by

$f(x) = \frac{1}{9} xe^{-\frac{x}{3} }$

The expected electric consumption is given by

= $\int\limits^a_0 {x \frac{1}{9} xe^{-\frac{1}{3} } } \, dx$

= $\frac{1}{9} \int\limits^a_0 { x^{2} e^{-\frac{1}{3} } } \, dx$

By integrating the given term for a= infinity we get,

expected consumption = 6

For there to be a shortage, the actual consumption must exceed the expected consumption. Thus, we need to find the P(consumption > 6)

⇒ P ( x > 6) = $\int\limits^a_6 {x \frac{1}{9} xe^{-\frac{1}{3} } } \, dx$

after integrating the term we get, P ( x > 6) = 0.406

Thus, the probability that there is power cut (shortage) on any given day is 0.406.

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